Question

Кабина лифта поднимается в течение первых 4с равноускоренно, достигая скорость 4 м/с. С этой скоростью она движется 8с, а затем 3с - равнозамедленно до остановки. Определите уть кабины лифта за всё время движения.

Answer 1

1) найдем ускорение в 1 случае
v = v0 + a*t
4 = 0 + a*4
a = 1 м / с^2
S1 = a * t * t / 2 = 1 * 4 * 4 / 2 = 8 м
2)
S2 = v * t = 4 * 8 = 32 м
3)
найдем ускорение в 3 случае
v = v0 + a * t
0 = 4 + a * 3
a = -4/3 м / с
S3 = v * t + a * t * t / 2 = 4 * 3 - 4 / 3 * 3 * 3 / 2 = 12 - 6 = 6 м
S = S1 + S2 + S3 = 6 + 8 + 32 = 46 м

Answer 2

Given Information:

The elevator cabin accelerates uniformly for the first 4 seconds, reaching a speed of 4 m/s. It then moves at this speed for 8 seconds, and finally decelerates uniformly for 3 seconds until it comes to a stop.

Calculation:

To find the total distance traveled by the elevator cabin, we need to calculate the distance traveled during each phase of the motion and then sum them up.

1. Distance traveled during acceleration: - The initial velocity, u, is 0 m/s. - The final velocity, v, is 4 m/s. - The time, t, is 4 seconds. - Using the formula distance = (initial velocity + final velocity) * time / 2, we can calculate the distance traveled during acceleration. - Plugging in the values, we get distance = (0 + 4) * 4 / 2 = 8 meters.

2. Distance traveled at constant speed: - The velocity, v, is 4 m/s. - The time, t, is 8 seconds. - Using the formula distance = velocity * time, we can calculate the distance traveled at constant speed. - Plugging in the values, we get distance = 4 * 8 = 32 meters.

3. Distance traveled during deceleration: - The initial velocity, u, is 4 m/s. - The final velocity, v, is 0 m/s. - The time, t, is 3 seconds. - Using the formula distance = (initial velocity + final velocity) * time / 2, we can calculate the distance traveled during deceleration. - Plugging in the values, we get distance = (4 + 0) * 3 / 2 = 6 meters.

4. Total distance traveled: - Summing up the distances from each phase, we get total distance = 8 + 32 + 6 = 46 meters.

Answer:

The elevator cabin travels a total distance of 46 meters during the entire duration of its motion.

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